Ask question

Ask question

All categories

2 years ago

14

If the child pulls with a force of 20 N for 12.0 m , and the handle of the wagon is inclined at an angle of 25 ∘ above the horiz

ontal, how much work does the child do on the wagon?

1 answer:

lilavasa [31]2 years ago

4 0

Answer: 217.52 N

Explanation: The applied force is 20 N, the distance covered is 12.0 m and the angle is 25° above the horizontal.

Hence the formulae that defines work done is given by

W = Force × distance

But since the force has been inclined at an angle θ above the horizontal, the horizontal component of force is neccesary to produce the required motion to make the child do work on the wagon.

Hence

Work done = (horizontal component of force) × distance

Work done = F cos θ × distance

Work done = 20 cos 25 × 12 = 217.52 N

You might be interested in

A car collides into a concrete wall going 25.0 m/s . It stops in 0.141 seconds and has a change in momentum of 39,400. What is t

vodka [1.7K]

Answer:

Mass of the car is 1576 kg.

Explanation:

Let the mass of the car be $m$ kg.

Given:

Initial velocity of the car is, $u=25\ m/s$

As the car stops, final velocity of the car is, $v=0\ m/s$

Change in momentum is, $\Delta p=39400$

Now, we know that, momentum is given as the product of mass and velocity.

So, change in momentum is given as:

$\Delta p=m(u-v)\\39400=m(25-0)\\39400=25m\\m=\frac{39400}{25}\\m=1576\ kg$

Therefore, the mass of the car is 1576 kg.

4 0

2 years ago

What is the magnitude of the velocity when the elastic potential energy is equal to the kinetic energy? (Assume that U=0 at equi

zhannawk [14.2K]

Answer:

Explanation:

General Equation of SHM is given by

$x=A\cos \omega t$

$v=-A\omega \sin \omega t$

where x=position of particle

A=maximum Amplitude

$\omega =$ angular frequency

t=time

At any time Total Energy is the sum of kinetic Energy and Elastic potential Energy i.e. $\frac{1}{2}kA^2$

where k=spring constant

Potential Energy is given by $U=\frac{1}{2}kx^2$

also it is given that Potential Energy(U) is equal to Kinetic Energy(K)

Total Energy $=K+U$

Total $=2U=2\times \frac{1}{2}kx^2$

$\frac{1}{2}kA^2=2\times \frac{1}{2}kx^2$

$x=\pm \frac{A}{\sqrt{2}}$

at $x=\frac{A}{\sqrt{2}}$

velocity is $v=\frac{A\omega}{\sqrt{2}}$

6 0

2 years ago

Charge q is accelerated starting from rest up to speed v through the potential difference V. What speed will charge q have after

adoni [48]

Answer: v = $1.19 * 10^{6} m/s$

Explanation: q = magnitude of electronic charge = $1.609 * 10^{-19} c$

mass of an electronic charge = $9.10 * 10^{-31} kg$

V= potential difference = 4V

v = velocity of electron

by using the work- energy theorem which states that the kinetic energy of the the electron must equal the work done use in accelerating the electron.

kinetic energy = $\frac{mv^{2} }{2}$ , potential energy = qV

hence, $\frac{mv^{2} }{2} = qV$

$\frac{9.10 *10^{-31} * v^{2} }{2} = 1.609 * 10^{-16} * 4\\\\\\\\9.10*10^{-31} * v^{2} = 2 * 1.609 *10^{-16} * 4\\\\\\9.10 *10^{-31} * v^{2} = 1.287 *10^{-15} \\\\v^{2} = \frac{1.287 *10^{-15} }{9.10 *10^{31} } \\\\v^{2} = 1.414*10^{15} \\\\v = \sqrt{1.414*10^{15} } \\\\v = 1.19 * 10^{6} m/s$

7 0

2 years ago

You skip north for 12 minutes to your best friend's house that is 1.5 kilometers away. What is your average velocity?

bagirrra123 [75]

Answer:

The average velocity is 7.5 km/h

Explanation:

Let's convert minutes to hours so our answer can be given in a common units of km/hour:

12 minutes = 12/60 hours = 0.2 hours

Now we estimate the average velocity calculating the distance travelled over the time it took:

1.5 / 0.2 km/h = 7.5 km/h

3 0

3 years ago

An accepted value for the acceleration due to gravity is 9.801 m/s2. In an experiment with pendulums, you calculate that the val

Fed [463]

g Generally the accepted value of acceleration due to gravity is 9.801 $m/s^2$

as per the question the acceleration due to gravity is found to be 9.42 $m/s^2$ in an experiment performed.

the difference between the ideal and observed value is 0.381.

hence the error is - $\frac{0.381}{9.801} *100$

=3.88735 percent

the error is not so high,so it can be accepted.

now we have to know why this occurs-the equation of time period of the simple pendulum is give as- $T=2\pi\sqrt[2]{l/g}$

$g=4\pi^2\frac{l}{T^2}$

As the experiment is done under air resistance,so it will affect to the time period.hence the time period will be more which in turn decreases the value of g.

if this experiment is done in a environment of zero air resistance,we will get the value of g which must be approximately equal to 9.801 $m/s^2$

5 0

3 years ago